Five points on $(x-y)$ plane $ (x1,y1), (x2,y2),... (x5,y5)$ defining a common conic intersection curve $C$ are obtained by section with variably inclined right circular cones i.e.,variable vertex $V$ position.

Same Intrxn of Cones

Find locus $L$ of all vertices $V.$


We have $ \epsilon$= eccentricity , $\gamma=$ semi-vertical angle of cone, $\alpha =$ angle to symmetry axis and since


the above relation implies that that every conic section is associated uniquely with an intersected right circular cone upto

  • scale
  • rotation around axis of symmetry and
  • translation

Special case : If $C$ is a circle then $L$ is the common axis of symmetry of each cone.

Thanks in advance for pointers to vertex locus finding.

  • $\begingroup$ I vaguely recall seeing this problem described in one of the old geometry books. I think it was in Somerville’s book. $\endgroup$
    – bubba
    Jan 23, 2022 at 8:22

1 Answer 1


If the given conic is either an ellipse or a hyperbola, vertex $V$ of the cone must lie on the plane perpendicular to the plane of the conic and passing through its foci.

Let $A$ and $B$ be the endpoints of the transverse axis of the conic, $V$ the cone vertex, and $2u$ the aperture angle of the cone. Let then $P$ be any point on the ellipse and $H$ its projection onto $AB$. A plane through $P$, perpendicular to the axis of the cone, intersects the cone along a circle $A'B'P$ (see diagram below), where $A'$ and $B'$ lie on plane $VAB$.

By the intersecting chords theorem we know that $PH^2=A'H\cdot B'H$. But, on the other hand, we have by similitude: $$ A'H:AH=BC:AB, \quad\hbox{that is:}\quad A'H={BC\over AB}\cdot AH; $$ $$ B'H:BH=AD:AB, \quad\hbox{that is:}\quad B'H={AD\over AB}\cdot BH. $$

If we set $n=VA$, $m=VB$ and $2a=AB$, the above formulas can be written as $$ A'H={m\sin u\over a}\cdot AH, \quad B'H={n\sin u\over a}\cdot BH, $$ and inserting these into the formula for $PH^2$ we get: $$ PH^2={mn\sin^2 u\over a^2}\,AH\cdot BH. $$ If $H$ is a focus of the conic, then $PH={b^2/a}$ is the semi-latus rectum, while $AH\cdot BH=|a^2-c^2|=b^2$ (as usual $2b$ is the length of the conjugate axis and $2c$ is the distance between foci) and from the above equation we get: $$ \tag{1} mn={b^2\over \sin^2 u}. $$ Another equation for $m$ and $n$ can be found from the cosine rule applied to triangle $AVB$: $$ \tag{2} m^2+n^2\mp2mn\cos2u=4a^2, $$ where sign $-$ must be taken for an ellipse ($\angle AVB=2u$) and sign $+$ for a hyperbola ($\angle AVB=\pi-2u$).

enter image description here

If the conic is an ellipse one obtains from $(1)$ and $(2)$ $$(m-n)^2=4(a^2-b^2),$$ that is the difference of the distances of $V$ from $A$ and $B$ is constant. The locus of $V$ is then a hyperbola, having its foci at the vertices of the ellipse and semi-major axis $\sqrt{a^2-b^2}$. This hyperbola thus passes through the foci of the ellipse.

If the conic is a hyperbola you get instead $$(m+n)^2=4(a^2+b^2)$$ and the locus of $V$ is an ellipse having its foci at the vertices of the hyperbola and semi-major axis $\sqrt{a^2+b^2}$. This ellipse thus passes through the foci of the hyperbola.

Finally, if the conic is a parabola of vertex $A$ and semi-latus rectum $p$ one gets from a similar reasoning: $$ AV=n={p\over2\sin^2u}. $$ From that it follows that the distance of vertex $V$ from the plane of the parabola is $$VK=n\sin2u=p\cot u,$$ while the distance from the projection of $V$ on that plane to the focus of the parabola is $$KF=n\cos2u+p/2={p\over2}\cot^2u.$$ Hence $KF=VK^2/(2p)$ and the vertex thus lies on a a parabola, equal to the given parabola and having its vertex at the focus of the latter.

We can summarise the above results as follows:

the locus of $V$ is a conic section, having center and transverse axis in common with the given conic but lying in an orthogonal plane, with eccentricity reciprocal to that of the given conic and vertices at the foci of the given conic.

You can see below an example: the green ellipse on plane x-y has vertices $A$, $B$, foci $F$, $G$ and eccentricity $1/\sqrt2$. The locus of the vertex $V_1$ of any cone intercepting the ellipse is the pink hyperbola on plane x-z, with vertices $F$, $G$, foci $A$, $B$ and eccentricity $\sqrt2$. Conversely, the ellipse is the locus of the vertex $V_2$ of any cone intercepting the hyperbola. Note also that the axes of the cones are tangent to the locus where the vertex lies.

enter image description here

  • $\begingroup$ Thank you. It will be very nice to be able to see Geogebra 3D images e.g. with eccentricities (section, vertex locus) = $(\sqrt 2, 1/\sqrt 2),\,(1/\sqrt 2,\sqrt 2)$ $\endgroup$
    – Narasimham
    May 30, 2019 at 23:04
  • $\begingroup$ I'll see if I can recover the GeoGebra file I used for that old answer and modify it. $\endgroup$ May 31, 2019 at 8:33
  • $\begingroup$ Hi, can you help me here? math.stackexchange.com/questions/4182690/… $\endgroup$
    – nonuser
    Jun 25, 2021 at 10:45

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